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Question: One way of writing the equation of state for heat gases is \[PV=RT\left[ 1+\dfrac{B}{V}+...... \righ...

One way of writing the equation of state for heat gases is PV=RT[1+BV+......]PV=RT\left[ 1+\dfrac{B}{V}+...... \right] where B is constant. An approximate expression for ’ B’ in terms of van der Waals constant ‘a’ and ‘b’ is:
A. b+a/RTb+a/RT
B. ba/RTb-a/RT
C. b+a/2RTb+a/2RT
D. a+b/RTa+b/RT

Explanation

Solution

According to the ideal gas law, the molecules of a gas are point particles with perfectly elastic collisions since gas particles aren’t point particles, a modification in the ideal gas law was done.
Ideal gas law before modification: PV=nRTPV=nRT and after modification
(P+aV2)(Vb)=RT\left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT
Where a, b are constants; P = pressure
V = Volume, T = Temp and R = constant

Complete step by step answer:
According to the modified Van Der Wall’s equation, we have
(P+aV2)(Vb)=RT\Rightarrow \left( P+\dfrac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT
P+aV2=RTVb\Rightarrow P+\dfrac{a}{{{V}^{2}}}=\dfrac{RT}{V-b}
P=RTVbαV2\Rightarrow P=\dfrac{RT}{V-b}-\dfrac{\alpha }{{{V}^{2}}}
PRT=1VbaV2RT\Rightarrow \dfrac{P}{RT}=\dfrac{1}{V-b}-\dfrac{a}{{{V}^{2}}RT}
VRT=1VbVaVRT\Rightarrow \dfrac{V}{RT}=\dfrac{1}{\dfrac{V-b}{V}}-\dfrac{a}{VRT}
PVRT=11bVaVRT......................(i) \Rightarrow \dfrac{PV}{RT}=\dfrac{1}{1-\dfrac{b}{V}}-\dfrac{a}{VRT}......................\text{(i) }

We know the binomial expansion:
(11x)=1+x+x2+x3+.......\left( \dfrac{1}{1-x} \right)=1+x+{{x}^{2}}+{{x}^{3}}+.......

Expanding equation 1 with the same binomial expansion we get,
RVRT=(1+bV+b2V2+.........)aVRT\Rightarrow \dfrac{RV}{RT}=\left( 1+\dfrac{b}{V}+\dfrac{{{b}^{2}}}{{{V}^{2}}}+......... \right)-\dfrac{a}{VRT}
PVRT=1+(baRT)1V+b2V2+......... (ii)\Rightarrow \dfrac{PV}{RT}=1+\left( b-\dfrac{a}{RT} \right)\dfrac{1}{V}+\dfrac{{{b}^{2}}}{{{V}^{2}}}+.........\text{ }\left( ii \right)
Comparing equation 2 with the equation give in this question we get,
B=baRTB=b-\dfrac{a}{RT}
At fixed temperature the Vander Waals equation describes the relation between pressure (P), volume (V) and Temperature (T).
So, the correct answer is “Option B”.

Note: 1. At a critical temperature, the gas is characterized by the critical values of Tk,Pk{{T}_{k}},{{P}_{k}} and Vk{{V}_{k}} which are determined only by the gas properties.
2. Vander Waals define the physical state of a homogeneous gas, is a modification of the ideal gas equation and moves nearly describes the properties of actual glass.
3. The Vander Waals equation corrects the ideal gas law for two major points. That is
(a) Excluded volume of gas particles.
(b) Alternative forces between gas molecules .